package cn.icatw.leetcode.editor.cn;
// 给你一个由 '1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。
//
// 岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。
//
// 此外，你可以假设该网格的四条边均被水包围。
//
//
//
// 示例 1：
//
//
// 输入：grid = [
//  ["1","1","1","1","0"],
//  ["1","1","0","1","0"],
//  ["1","1","0","0","0"],
//  ["0","0","0","0","0"]
//]
// 输出：1
//
//
// 示例 2：
//
//
// 输入：grid = [
//  ["1","1","0","0","0"],
//  ["1","1","0","0","0"],
//  ["0","0","1","0","0"],
//  ["0","0","0","1","1"]
//]
// 输出：3
//
//
//
//
// 提示：
//
//
// m == grid.length
// n == grid[i].length
// 1 <= m, n <= 300
// grid[i][j] 的值为 '0' 或 '1'
//
//
// Related Topics 深度优先搜索 广度优先搜索 并查集 数组 矩阵 👍 2557 👎 0


// Java：岛屿数量
public class T200_NumberOfIslands {
    public static void main(String[] args) {
        Solution solution = new T200_NumberOfIslands().new Solution();
        // TO TEST
    }

    // leetcode submit region begin(Prohibit modification and deletion)
    class Solution {
        public int numIslands(char[][] grid) {
            if (grid == null || grid.length == 0) {
                return 0;
            }
            // int m = grid.length;
            int count = 0;
            for (int i = 0; i < grid.length; i++) {
                for (int j = 0; j < grid[i].length; j++) {
                    if (grid[i][j] == '1') {
                        count++;
                        dfs(grid, i, j);
                    }
                }
            }
            return count;
        }

        private void dfs(char[][] grid, int i, int j) {
            if (i < 0 || j < 0 || i >= grid.length || j >= grid[0].length || grid[i][j] == '0') {
                return;
            }
            grid[i][j] = '0';
            dfs(grid, i + 1, j);
            dfs(grid, i - 1, j);
            dfs(grid, i, j + 1);
            dfs(grid, i, j - 1);
        }
    }
// leetcode submit region end(Prohibit modification and deletion)

}
